Question

How do you find the indefinite integral of `int (x^2-3x^2+5)/(x-3)`?

Answer 1

Start by simplifying the numerator.

`=>int(5 - 2x^2)/(x - 3)`

We now divide `5 - 2x^2` by `x- 3`.

Hence, our integral becomes `int(-2x - 6 - 13/(x - 3))`, which can be integrated using the rule `int(x^n)dx = x^(n + 1)/(n + 1) + C` and the rule `int(1/u)du = ln|u| + C`.

`=>-x^2 - 6x - 13ln|x - 3| + C`

Hopefully this helps!